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Comparisons of stochastic task-resource systems

Bruno Gaujal Jean-Marc Vincent

INRIA and LIG

Fréjus, 4 Juin 2007

B. Gaujal, J.-M. Vincent (UJF, INRIA ) Stochastic task-resource systems Fréjus, 4 Juin 2007 1 / 30

Outline

1 Probabilistic task-resource models

2 Stochastic orders

3 Comparison of systemsMapping techniqueassociationCoupling technique

4 Several applicationsPERT GraphQueuespolling from several queues

Probabilistic models

The main object of this lecture is the task-resource model.

Tasks are characterized by the arrival times that form a point process 0 6 T1 6 T2 6 T3 6 · · ·over the positive real line and by the sizes of tasks that is a sequence of real numbers S1, S2, . . ..

Resources are characterized by their number K ∈ N ∪ +∞ and the respective speedsv1, . . . , vK .

In the following the size of the tasks is often given is seconds (time for a resource of speed 1 totreat a task).

Additionally, tasks and ressources may be constrainted by dependencies, synchronizations,availability conditions, matchings, . . .

Here we will mostly consider very simple systems with one common features : randomness.Basically, the arrival times and/or the task sizes will be random processes.

The main object of this lecture is the task-resource model.Tasks are characterized by the arrival times that form a point process 0 6 T1 6 T2 6 T3 6 · · ·over the positive real line and by the sizes of tasks that is a sequence of real numbers S1, S2, . . ..

Stochastic Orders

There are many ways to compare two random variables (and random processes). The mostobvious one is to compare the means : X 6µ Y if E(X) 6 E(Y ).

However, this order is rather crude and may not capture a lot of insight in the comparison of twostochastic systems, in particular in the scheduling context.

Consider a two task-resource systems with one resource. One has arrivals every 4 seconds oftasks of size 4 and the other has arrivals of tasks of size 2 at times 5n and 5n + 1.Then the expected input loads per second are E(L1) = 1 > E(L2) = 4/5 respectively.As for the expected waiting times, E(W1) = 0 < E(W2) = 1/2.Even if waiting times are increasing functions of loads (see later), this is not the case for the µorder.

There exists several stochastic orders (in the book Comparison methods for stochastic models andrisks (Muller and Stoyan, 2002), 49 different orders are defined with different applications in mind.

Stochastic Orders

Consider a two task-resource systems with one resource. One has arrivals every 4 seconds oftasks of size 4 and the other has arrivals of tasks of size 2 at times 5n and 5n + 1.

Then the expected input loads per second are E(L1) = 1 > E(L2) = 4/5 respectively.As for the expected waiting times, E(W1) = 0 < E(W2) = 1/2.Even if waiting times are increasing functions of loads (see later), this is not the case for the µorder.

Stochastic Orders

Consider a two task-resource systems with one resource. One has arrivals every 4 seconds oftasks of size 4 and the other has arrivals of tasks of size 2 at times 5n and 5n + 1.Then the expected input loads per second are E(L1) = 1 > E(L2) = 4/5 respectively.

As for the expected waiting times, E(W1) = 0 < E(W2) = 1/2.Even if waiting times are increasing functions of loads (see later), this is not the case for the µorder.

Stochastic Orders

Consider a two task-resource systems with one resource. One has arrivals every 4 seconds oftasks of size 4 and the other has arrivals of tasks of size 2 at times 5n and 5n + 1.Then the expected input loads per second are E(L1) = 1 > E(L2) = 4/5 respectively.As for the expected waiting times, E(W1) = 0 < E(W2) = 1/2.

Even if waiting times are increasing functions of loads (see later), this is not the case for the µorder.

Stochastic Orders

The usual stochastic order

The usual stochastic order (also called the strong order) is defined as follows (for real randomvariables).X 6st Y if FX (a) = P(X 6 a) > P(Y 6 a) = FY (a) for all a.

The st order has several other characterizations :

Sample path definition There exists two variables X ′ and Y ′ in (Ω, A, P) with the samedistribution as X and Y such that X(ω) 6 Y (ω) for each ω ∈ Ω.

Integral definition For all increasing function f , E(f (X)) 6 E(f (Y )).

The usual stochastic order

The usual stochastic order (also called the strong order) is defined as follows (for real randomvariables).X 6st Y if FX (a) = P(X 6 a) > P(Y 6 a) = FY (a) for all a.

The st order has several other characterizations :

Sample path definition There exists two variables X ′ and Y ′ in (Ω, A, P) with the samedistribution as X and Y such that X(ω) 6 Y (ω) for each ω ∈ Ω.

Integral definition For all increasing function f , E(f (X)) 6 E(f (Y )).

The usual stochastic order : examples

1. Show that X 6st Y ⇒ X 6µ Y .

2. Show that X 6st Y and EX = EY ⇒ FX = FY .

3. Compare the following integer random variables :

X = 1 w.p. 1/4, 2 w.p. 1/2, 3 w,.p. 1/4Y = 2 w.p. 1/2 , 3 w.p. 1/2Z = 1 w.p. 1/3 , 3 w.p. 2/3

Stronger stochastic ordersSome orders are stronger than st :Consider the following case : somebody wants to buy a car and can choose between two modelswith lifetimes X and Y . If the price is the same and X 6st Y them, she ought to buy model Y . Nowwhat happens if both cars are used (one year old), is Y still a better choice ?

well, not necessarily :Assume that X is uniform over [0, 3] (with db F ) and Y has a distribution with density 1/6, 1/2, 1/3on [0, 1], ]1, 2], ]2, 3] (with db G).Then, X 6st Y (F > G). However X1 = (X |X > 1) and Y1 = (Y |Y > 1) are not st-comparable :X1 is uniform over [0, 2] (with density 1/2) and Y1 has a distribution with density 3/5, 2/5, on[0, 1], ]1, 2].

G F−1

What order is preserved under aging ?

Stronger stochastic ordersSome orders are stronger than st :Consider the following case : somebody wants to buy a car and can choose between two modelswith lifetimes X and Y . If the price is the same and X 6st Y them, she ought to buy model Y . Nowwhat happens if both cars are used (one year old), is Y still a better choice ?well, not necessarily :

Assume that X is uniform over [0, 3] (with db F ) and Y has a distribution with density 1/6, 1/2, 1/3on [0, 1], ]1, 2], ]2, 3] (with db G).Then, X 6st Y (F > G). However X1 = (X |X > 1) and Y1 = (Y |Y > 1) are not st-comparable :X1 is uniform over [0, 2] (with density 1/2) and Y1 has a distribution with density 3/5, 2/5, on[0, 1], ]1, 2].

G F−1

Stronger stochastic ordersSome orders are stronger than st :Consider the following case : somebody wants to buy a car and can choose between two modelswith lifetimes X and Y . If the price is the same and X 6st Y them, she ought to buy model Y . Nowwhat happens if both cars are used (one year old), is Y still a better choice ?well, not necessarily :Assume that X is uniform over [0, 3] (with db F ) and Y has a distribution with density 1/6, 1/2, 1/3on [0, 1], ]1, 2], ]2, 3] (with db G).

Then, X 6st Y (F > G). However X1 = (X |X > 1) and Y1 = (Y |Y > 1) are not st-comparable :X1 is uniform over [0, 2] (with density 1/2) and Y1 has a distribution with density 3/5, 2/5, on[0, 1], ]1, 2].

G F−1

Stronger stochastic ordersSome orders are stronger than st :Consider the following case : somebody wants to buy a car and can choose between two modelswith lifetimes X and Y . If the price is the same and X 6st Y them, she ought to buy model Y . Nowwhat happens if both cars are used (one year old), is Y still a better choice ?well, not necessarily :Assume that X is uniform over [0, 3] (with db F ) and Y has a distribution with density 1/6, 1/2, 1/3on [0, 1], ]1, 2], ]2, 3] (with db G).Then, X 6st Y (F > G). However X1 = (X |X > 1) and Y1 = (Y |Y > 1) are not st-comparable :X1 is uniform over [0, 2] (with density 1/2) and Y1 has a distribution with density 3/5, 2/5, on[0, 1], ]1, 2].

G F−1

What order is preserved under aging ?B. Gaujal, J.-M. Vincent (UJF, INRIA ) Stochastic task-resource systems Fréjus, 4 Juin 2007 7 / 30

The hazard rate order : hr

The hazard rate (or failure rate) is defined by :

rX (t) = limε→0

P(X < t + ε|X > t)ε

=fX (t)

1− Fx (t)= −

ln(1− FX (t))

DefinitionX 6hr Y if rX (t) > rY (t).

the Proba-Proba plot G(F−1(t)) is star shaped with respect to (1,1).

Eg(X∗, Y∗) 6 Eg(Y∗, X∗) ∀g s.t. g(x , y)− g(y , x) increasing in x , ∀x > y .

The hr order is preserved under aging and is stronger than the st order

rX (t) = limε→0

P(X < t + ε|X > t)ε

=fX (t)

1− Fx (t)= −

ln(1− FX (t))

rX (t) = limε→0

P(X < t + ε|X > t)ε

=fX (t)

1− Fx (t)= −

ln(1− FX (t))

rX (t) = limε→0

P(X < t + ε|X > t)ε

=fX (t)

1− Fx (t)= −

ln(1− FX (t))

The likelihood order : lr

Another order which is even stronger than hr is the likelihood ratio which preserves st under anyconditioning :

DefinitionU = [a, b], V = [c, d ], U < V X 6lr Y if P(X ∈ V )P(Y ∈ U) 6 P(X ∈ U)P(Y ∈ V ) orequivalently (X |X ∈ U) 6st (Y |Y ∈ U)

The P-P plot is convex.

Eg(X∗, Y∗) 6 Eg(Y∗, X∗) ∀g s.t. g(x , y)− g(y , x) > 0, ∀x > y .

The lr order is preserved under any conditioning and is stronger than the hr order.

Other stochastic orders : convex orders

The convex orders are used to compare the variability of stochastic variables.

DefinitionX 6cx Y if Ef (X) 6 Ef (Y ) for all convex functions f .

DefinitionX 6icx Y if Ef (X) 6 Ef (Y ) for all increasing convex functions f .

Strassen Representation Theorem :

TheoremX 6cx Y iff there exist two r.v. X ′ and Y ′ with the same db as X and Y such that X ′ = E(Y ′|X ′).

Corollary if X and Z are independent and E(Z ) = 0 then, X 6cx X + Z .

Discrete dynamical systems

There are two types of models in scheduling.

static models :X = φ(Z1, · · · , ZN) anddynamic models Xn = φn(Xn−1, Zn), ∀n > 0.

A dynamical system is time -monotone for order F if Xn 6F Xn−1.A system (static or dynamic) is F- isotone if Zk 6F Z ′k ⇒ Xn 6F X ′n.

Comparison are often proved using mapping, coupling, association and monotony.

Mapping techniques

Principle : Prove comparability by comparing the inputs of functionals.for a static system,

Theoremif (Z1, . . . , Zn) 6st (Z ′1, . . . Z ′n) and are independent, then if Φ is increasing , thenΦ(Z1, . . . , Zn) 6st Φ(Z ′1, . . . Z ′n).if (Z1, . . . , Zn) 6icx (Z ′1, . . . Z ′n) and are independent, then If Φ is increasing and convex thenΦ(Z1, . . . , Zn) 6icx Φ(Z ′1, . . . Z ′n).

For a dynamic system,

Theoremif (Z1, . . . , Zn) 6F (Z ′1, . . . Z ′n) and all ϕn are increasing, (resp. increasing and convex) thenXn 6F X ′n, with F = st (resp. F = icx).

Association

Two random variables X and Y are associated if cov(g(X), f (Y )) > 0 for all increasing f and g.

Xn = φ(Xn, Zn) if φ is monotone on both variables, then if Zn are independent or associated thenXn are associated.

Finally, (X1, . . . Xn) associated implies that (X1, . . . Xn) 6o (X∗1 , . . . X∗n ) where (X∗1 , . . . X∗n ) areindependent versions of (X1, . . . Xn).

Association

Coupling technique

Principle : Prove comparisons using sample paths. Coupling often provides more powerful results.

Example : compare the load C for two task-resource systems with s and s′ resources,respectively :How to show that Cn 6st C′n ? (loads at the nth arrival, Tn)

The best is by using a coupling method, i.e. by considering a unique input process sample ω andcompare the two systems over that single sequence of sizes and arrival times. The workloadvectors Cn and C′n are the increasingly ordered workloads at time Tn in the different resources(they have s and s′ components respectively). One has Cn = R(Cn−1 + Sne1 − δn1)+ Then, itshould be clear that the s first components of Cn and C′n are comparable, by induction. Therefore,Cn =st Cn(1) 6st C′n(1) =st C′n.

Coupling technique

The best is by using a coupling method, i.e. by considering a unique input process sample ω andcompare the two systems over that single sequence of sizes and arrival times. The workloadvectors Cn and C′n are the increasingly ordered workloads at time Tn in the different resources(they have s and s′ components respectively).

One has Cn = R(Cn−1 + Sne1 − δn1)+ Then, itshould be clear that the s first components of Cn and C′n are comparable, by induction. Therefore,Cn =st Cn(1) 6st C′n(1) =st C′n.

Coupling technique

The best is by using a coupling method, i.e. by considering a unique input process sample ω andcompare the two systems over that single sequence of sizes and arrival times. The workloadvectors Cn and C′n are the increasingly ordered workloads at time Tn in the different resources(they have s and s′ components respectively). One has Cn = R(Cn−1 + Sne1 − δn1)+

Then, itshould be clear that the s first components of Cn and C′n are comparable, by induction. Therefore,Cn =st Cn(1) 6st C′n(1) =st C′n.

Coupling technique

The best is by using a coupling method, i.e. by considering a unique input process sample ω andcompare the two systems over that single sequence of sizes and arrival times. The workloadvectors Cn and C′n are the increasingly ordered workloads at time Tn in the different resources(they have s and s′ components respectively). One has Cn = R(Cn−1 + Sne1 − δn1)+ Then, itshould be clear that the s first components of Cn and C′n are comparable, by induction.

Therefore,Cn =st Cn(1) 6st C′n(1) =st C′n.

Coupling technique

G/G/2 G/G/1

Coupling technique

G/G/2 G/G/1

Coupling technique

G/G/2 G/G/1

Coupling technique

G/G/2 G/G/1

Coupling technique

G/G/2 G/G/1

The problem 1||∑

This is one of the simplest scheduling problem : one resource with no scheduling restriction on Ntasks, all arriving at time 0, while the objective is to minimize the sum of the completion times (orthe average completion time)We consider all tasks to be independent of sizes S1, . . . , SN .

For a given schedule (or permutation) σ, the objective function isTσ =

PNi=1 Ci =

PNi=1(N − i + 1)Sσ(i).

We consider two particular schedules : SEPT (Shortest Expected Processing Time) and LEPT(Largest Expected Processing Time).

It should be clear that for any permutation σ, ETSEPT 6 ETσ 6 ETLEPT .Indeed, ETσ = E

PNi=1 Ci =

PNi=1(N − i + 1)ESσ(i).

But can we say more ?

The problem 1||∑

PNi=1 Ci =

PNi=1(N − i + 1)Sσ(i).

PNi=1 Ci =

PNi=1(N − i + 1)ESσ(i).

The problem 1||∑

PNi=1 Ci =

PNi=1(N − i + 1)Sσ(i).

PNi=1 Ci =

PNi=1(N − i + 1)ESσ(i).

The problem 1||∑

PNi=1 Ci =

PNi=1(N − i + 1)Sσ(i).

PNi=1 Ci =

PNi=1(N − i + 1)ESσ(i).

The problem 1||∑

Ci revisited

Theorem (Shanthikumar, Yao, 1993)If Si 6lr Si+1 for all i , then TSEPT 6st Tσ 6st TLEPT .If Si 6hr Si+1 for all i , then TSEPT 6icx Tσ 6icx TLEPT .

Proof uses a classical interchange argument (done for hr)

For all σ 6= SEPT there exists k such that σ(k) = j, σ(k + 1) = i with i < j . Let µ = σ exceptµ(k) = i, µ(k + 1) = j (µ is closer to SEPT than σ).

Now, Tσ = Xj + k(Xi + Xj ) + Y and Tµ = Xi + k(Xi + Xj ) + Y where Y is the contribution of theother jobs, independent of Si and Sj .

Moreover g(x , y) = f (x + k(x + y)) satisfies g(x , y)− g(y , x) is increasing as long as f is convexand increasing. Therefore, Si 6hr Sj implies Ef (Xj + k(Xi + Xj )) > Ef (Xi + k(Xi + Xj )) for allincreasing convex f .

Finally, Xi + k(Xi + Xj ) 6icx Xj + k(Xi + Xj ) implies Tµ 6icx Tσ .

The problem 1||∑

Ci revisited

The problem 1||∑

Ci revisited

The problem 1||∑

Ci revisited

The problem 1||∑

Ci revisited

PERT Graph

A PERT graph is a more general static model : N tasks are to be executed over an infinite numberof resources and are constrained by an acyclic graph.

PERT graphs are impossible to solve (compute the makespan) analytically in general(Kamburowski, 1992). However, one can use comparisons to prove several results.

PERT Graph

A PERT graph is a more general static model : N tasks are to be executed over an infinite numberof resources and are constrained by an acyclic graph.

PERT graphs are impossible to solve (compute the makespan) analytically in general(Kamburowski, 1992). However, one can use comparisons to prove several results.

PERT Graph, continued

Here are the ingredients used to compare (and compute bounds) for PERT graphs.

1- Using the mapping technique :C = Φ(X1, . . . XN) = maxc∈P(G)

Pi∈c Xi .

Note that Φ is convex and increasing. This implies the following first result. Zi 6F Z ′i impliesC 6F C′ with F = st or icx .

2- Next, if tasks are independent (or associated), then the paths are all associated and aretherefore bounded by independent versions :for all c, Sc 6st S∗c and C 6st C∗ = maxc∈P(G) S∗c

3- Next, if Xi is NBUE (New Better than Used in Expectation : E(X − t |X > t) 6 EX ) thenXi 6cx exp(E(Xi )).

Pi∈c Xi .

2- Next, if tasks are independent (or associated), then the paths are all associated and aretherefore bounded by independent versions :

for all c, Sc 6st S∗c and C 6st C∗ = maxc∈P(G) S∗c

Pi∈c Xi .

This allows us to show that

maxc∈P(G)

Xi∈c

EXi 6icx C 6icx maxc∈P(G)

Xi∈c

exp(E(Xi ))

andC 6icx max

c∈P(G)exp(

Xi∈c

E(Xi )) =dbY

c∈P(G)

1− exp(−tXi∈c

E(Xi )),

as soon as Xi are all NBUE and associated.

This allows us to show that

maxc∈P(G)

Xi∈c

EXi 6icx C 6icx maxc∈P(G)

Xi∈c

exp(E(Xi ))

andC 6icx max

c∈P(G)exp(

Xi∈c

E(Xi )) =dbY

c∈P(G)

1− exp(−tXi∈c

E(Xi )),

as soon as Xi are all NBUE and associated.

Queues

Queues are among simplest dynamic systems, but are still the source of many open problems.Tasks do not have any constraints, sizes and arrival times are often independent.

Queues

Lindley’s formula

δδ δ δ δ

d d d d2 3 4 5d1

a a a a a1 2 3 4 5 a6

32 4 5

Wn is the waiting time of the n-th task. It is a dynamical system of the form Wn = ϕ(Wn−1, Xn)with Xn = Sn−1 − δn and ϕ defined by the Lindley’s equation :

Wn = max (Wn−1 + Xn, 0) .B. Gaujal, J.-M. Vincent (UJF, INRIA ) Stochastic task-resource systems Fréjus, 4 Juin 2007 21 / 30

Loynes’ scheme

TheoremWn 6st Wn+1 in a G/G/1 queue, initialy empty.

Proof. done by a backward coupling known as the Loynes’ scheme. Construct on a commonprobability space two trajectories by going backward in time : S1

i−n(ω) = S2i−n−1(ω) with

distribution Si and T 1i−n(ω) = T 2

i−n−1(ω), with distribution Ti − Tn+1 for all 0 6 i 6 n + 1 andS1−n−1(ω) = 0.

By construction, W 10 =st Wn and W 2

0 =st Wn+1. Also, it should be clear that0 = W 1

−n+1(ω) 6 W 2−n+1(ω) for all ω.

This implies W 1−i (ω) 6 W 2

−i (ω) so that Wn 6st Wn+1.

This has many consequences in terms of existence and uniqueness of a stationary (or limit)regime for the G/G/1 queue (Baccelli Bremaud, 2002).

Loynes’ schemeTheoremWn 6st Wn+1 in a G/G/1 queue, initialy empty.

−n+1(ω) 6 W 2−n+1(ω) for all ω.

0Ti − Tn+1−Tn+1

−n+1(ω) 6 W 2−n+1(ω) for all ω.

0Ti − Tn+1−Tn+1

−n+1(ω) 6 W 2−n+1(ω) for all ω.

0Ti − Tn+1−Tn+1

Input process

Folk theorem (Ross conjecture, 1978) :things work better when the input traffic has less variability.

Theoremif (W0, X1, . . . , Xn) 6F (W ′

0, X ′1, . . . , X ′n) then Wn 6F W ′n (with F = st or icx).

proof f (x , w) = max(w + x , 0) is convex and increasing for both variables.

Input process

Folk theorem (Ross conjecture, 1978) :things work better when the input traffic has less variability.

Theoremif (W0, X1, . . . , Xn) 6F (W ′

0, X ′1, . . . , X ′n) then Wn 6F W ′n (with F = st or icx).

proof f (x , w) = max(w + x , 0) is convex and increasing for both variables.

Examples

1 Show that if the traffic intensity is fixed in a single GI/GI/1 queue, then the average waitingtime is smallest when the arrivals are periodic.

2 Show that if the arrival process in a GI/M/1 queue is NBUE, then the average waiting time canbe bounded by EW 6 1

µ−1/E(T1).

Examples

1 Show that if the traffic intensity is fixed in a single GI/GI/1 queue, then the average waitingtime is smallest when the arrivals are periodic.

2 Show that if the arrival process in a GI/M/1 queue is NBUE, then the average waiting time canbe bounded by EW 6 1

µ−1/E(T1).

Extremal input processes

Several extensions are possible :

Theorem (Altman, Gaujal, Hordijk, 2003)If the arrival sequence T1, . . . , Tn, . . . is fixed in a stochastic FIFO event graph (arbitrary networkof queues with no branching enriched with fork and join nodes), then, S1, . . . , Sn 6cx S′1, . . . , S′nimplies (W1, . . . , Wn) 6icx (W1, . . . , Wn).

Service disciplineIn a queue with one or more servers, tasks may be served according to disciplines : FIFO, LIFO,PS, priority, random, . . .

PS has insensibility, reversibility and product form properties,FIFO (F) has optimality properties, in terms of waiting times :

TheoremIn a GI/GI/1 queue, f (WF

n ) 6st f (Wπn ) for all service discipline π and all convex increasing and

symmetric f .

Proof. using a coupling technique and majorization.Since service time and arrivals are independent, we can rearrange the service times in the orderof service (and not of arrivals) under policy π and F Let Di be the departure epochs. They coincideunder both policies.Then W F

i = Di − Ti and assume that π interchange the departure of j and j + 1 : W πj = Dj+1 − Tj

and W πj+1 = Dj − Tj+1

Now, it should be obvious that W πj + W π

j+1 = W Fj + W F

j+1 and if f is increasing convex andsymmetric (or Schur convex) f (W π

j , W πj+1) > f (W F

j , W Fj+1).

In general, consider all tasks (n) within a busy period of the system, then,W π

1 + · · ·+ W πj+1 = W F

1 + · · ·+ W Fn and interchanging a pair of customers out of order under π,

reduces the value of f (W π1 , . . . , W π

n ) down to the value of f (W F1 , . . . , W F

n ) for any Schur convexfunction f .If the first n tasks do not form a busy period, then W π

1 + · · ·+ W πj+1 > W F

1 + · · ·+ W Fn and again

f (W π1 , . . . , W π

n ) > f (W F1 , . . . , W F

n ) for any Schur convex function f

Service disciplineIn a queue with one or more servers, tasks may be served according to disciplines : FIFO, LIFO,PS, priority, random, . . .PS has insensibility, reversibility and product form properties,

FIFO (F) has optimality properties, in terms of waiting times :

symmetric f .

j+1 = W Fj + W F

j , W πj+1) > f (W F

j , W Fj+1).

1 + · · ·+ W πj+1 = W F

1 + · · ·+ W πj+1 > W F

f (W π1 , . . . , W π

n ) > f (W F1 , . . . , W F

Service disciplineIn a queue with one or more servers, tasks may be served according to disciplines : FIFO, LIFO,PS, priority, random, . . .PS has insensibility, reversibility and product form properties,FIFO (F) has optimality properties, in terms of waiting times :

symmetric f .

j+1 = W Fj + W F

j , W πj+1) > f (W F

j , W Fj+1).

1 + · · ·+ W πj+1 = W F

1 + · · ·+ W πj+1 > W F

f (W π1 , . . . , W π

n ) > f (W F1 , . . . , W F

symmetric f .

Proof. using a coupling technique and majorization.Since service time and arrivals are independent, we can rearrange the service times in the orderof service (and not of arrivals) under policy π and F Let Di be the departure epochs. They coincideunder both policies.

Then W Fi = Di − Ti and assume that π interchange the departure of j and j + 1 : W π

j = Dj+1 − Tj

j+1 = W Fj + W F

j , W πj+1) > f (W F

j , W Fj+1).

1 + · · ·+ W πj+1 = W F

1 + · · ·+ W πj+1 > W F

f (W π1 , . . . , W π

n ) > f (W F1 , . . . , W F

symmetric f .

j+1 = W Fj + W F

j , W πj+1) > f (W F

j , W Fj+1).

1 + · · ·+ W πj+1 = W F

1 + · · ·+ W πj+1 > W F

f (W π1 , . . . , W π

n ) > f (W F1 , . . . , W F

symmetric f .

j+1 = W Fj + W F

j , W πj+1) > f (W F

j , W Fj+1).

1 + · · ·+ W πj+1 = W F

1 + · · ·+ W πj+1 > W F

f (W π1 , . . . , W π

n ) > f (W F1 , . . . , W F

symmetric f .

j+1 = W Fj + W F

j , W πj+1) > f (W F

j , W Fj+1).

1 + · · ·+ W πj+1 = W F

n ) for any Schur convexfunction f .

If the first n tasks do not form a busy period, then W π1 + · · ·+ W π

j+1 > W F1 + · · ·+ W F

n and againf (W π

1 , . . . , W πn ) > f (W F

1 , . . . , W Fn ) for any Schur convex function f

symmetric f .

j+1 = W Fj + W F

j , W πj+1) > f (W F

j , W Fj+1).

1 + · · ·+ W πj+1 = W F

1 + · · ·+ W πj+1 > W F

f (W π1 , . . . , W π

n ) > f (W F1 , . . . , W F

Polling systems

Choosing the best open loop schedule for the server corresponds to choose the most regularservice in each queue. (Gaujal, Hordijk, Van der Laan, 2007)

Example : for two queues (1 and 2) 12121212. . . is a better schedule than 1122112211. . .The main difficulty is to compute the frequency of the visits to each queue.

Polling systems

Example : for two queues (1 and 2) 12121212. . . is a better schedule than 1122112211. . .

The main difficulty is to compute the frequency of the visits to each queue.

Polling systems

Polling systems, continued

αopt = 1/2

Instability

FIG.: The frequency of the server allocations w.r.t input intensities

Polling systems, continued

0.75 0.8 0.85 0.9 0.95 1

FIG.: The frequency of the server allocations w.r.t the total load, the ratio of input intensities being fixed.

Conclusion

There exists a systematic framework to deal with task-resource systems involving randomnessthrough the theory of stochastic comparisons.

Main actors in that field :R. Righter, Z. Liu, J . Shanthikumar, C. Cassandras, T. Rolski. . .

Main bibliography for that talk :Comparison Methods for Stochastic Models and Risks (A. Muller and D. Stoyan, 2002).Stochastic Modeling and the Theory of Queues (R. W. Wolff, 1989).Discrete-Event Control of Stochastic Networks : Multimodularity and Regularity (E. Altman, B.Gaujal and A. Hordijk, 2003).

Conclusion

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